If a square is colored red and blue, must there be either a red path connecting the top and bottom, or a blue path connecting the left and right? In the board game Hex, players take turns coloring hexagons either red or blue. One player tries to connect the top and bottom edges of the board, colored red; the other tries to connect the left and right edges, colored blue. It is known that a game of Hex will never end in a tie: no matter how it is played, there will always be either a blue path connecting the blue edges, or a red path connecting the red edges.
My question is, if this fact always holds for a finite grid of hexagons, does it also hold on the plane? If the top and bottom edges of a square are colored red, the left and right edges are colored blue, and the interior of the square is colored arbitrarily, must there be either a red path connecting the red edges, or a blue path connecting the blue edges?
More formally, let $S$ be any subset of $[0, 1]^2$. $S$ will represent the points that are red. Must there be either a path within $S$ whose endpoints are of the form $(x, 0)$ and $(x, 1)$, or a path within $[0, 1]^2 - S$ whose endpoints are of the form $(0, y)$ and $(1, y)$?
 A: An answer to this came to me today as I was thinking about how I might describe connectedness to a layperson. Path-connectedness is easy enough: a set is path-connected if, given two points in the set, you can get from one to the other by following a path within the set. If I'm not mistaken, connectedness in the plane can be defined like so: a set is connected if you can't draw a curve separating the set into two parts, such that the curve lies entirely outside the set.
So, if we have an example of a set that is connected but not path-connected, we ought to be able to make it into a counterexample. And sure enough, the topologist's sine curve seems to do the trick. Let $S$ be the graph of the function
$$f(x) =\frac12 + \frac14 \sin \left (\frac{1}{x - 1/2} \right), \text{ if $x \ne 1/2$}; \qquad 1/2, \text{ otherwise}.$$
Since this function is not continuous (and its graph is not path-connected), it is impossible to go from the left edge to the right edge while staying inside $S$. But since the function's graph is connected, it is impossible to go from the top edge to the bottom edge while staying outside the function—if this were possible, we would be able to separate the graph into the part to the left of the curve, and the part to the right of the curve, showing that the graph is disconnected.
A: If the path is required to be continuous, the answer is no. 
Since $[0,1]$ is separable, there are only $2^\omega$ continuous functions from $[0,1]$ into $[0,1]^2$; enumerate those that yield paths connecting opposite sides of the square as $\{\varphi_\xi:\xi < 2^\omega\}$. For $\xi < 2^\omega$ recursively choose points $p_\xi,q_\xi \in [0,1]^2$ as follows. Suppose that $\eta < 2^\omega$, and points $p_\xi$ and $q_\xi$, all distinct, have been chosen for all $\xi<\eta$. Clearly $$|\{p_\xi:\xi<\eta\}\cup\{q_\xi:\xi<\eta\}| < 2^\omega,$$ but $|\operatorname{ran}\varphi_\eta| = 2^\omega$, so we may choose distinct points $$p_\eta,q_\eta \in \operatorname{ran}\varphi_\eta \setminus \left(\{p_\xi:\xi<\eta\}\cup\{q_\xi:\xi<\eta\}\right),$$ and the construction goes through to $2^\omega$.
Now color the points of the set $\{p_\xi:\xi < 2^\omega\}$ blue and those of the set $\{q_\xi:\xi < 2^\omega\}$ red; any remaining points of $[0,1]$ may be colored either blue or red. Then every continuous path in the unit square passes through at least one point of each color.
A: It is even true that $I^{2}$ may be represented as an union of two closed sets
$S_{1}$ and $S_{2}$ none of them containing a topological arc connecting two
opposite sides of the square. Here is a possible example.
Let $D$ be a closed disk in the interior of $I^{2}$ , take 4 disjoint
topological spirals $J_{i}$ lying in $I^{2}$ and homeomorphic to $[0,\infty)$,
each of them beginning from a different side of the square and spiralling
around $D$ in such a way that $\overline{J_{i}}=J_{i}\cup\partial D$. Now
thicken slightly each $J_{i}$ to get some $J_{i}^{\prime}\supset J_{i}$
homeomorphic to a closed half-plane so that the $J_{i}^{\prime}$ are still
disjoint and spiralling around $D$. [$\overline{J_{i}^{\prime}}=J_{i}^{\prime
}\cup\partial D$.] To get the claimed representation of the square, it suffices
to set
 $S_{1}=(\cup_{i=1}^{4}J_{i}^{\prime})\cup D$ and $S_{2}=\overline
{I^{2}\backslash S_{1}}$. 
